統計力学演習問題(2)

統計力学演習問題 (2)
番号:
名前:
[1](1)(a)
(2)
∫
∫
F · dr =
C
F·
C
∫ 1
dr
dt
dt
∂Fy
∂Fx
−
= 2 − (−5y)
∂y
∂x
= 5y + 2 ̸= 0
(3t2 + 2t2 , −5t · t2 ) · (1, 2t)dt
=
0
[2](1)(a)
5
−2
3
1
=−
3
=
∫
∫
F · dr =
C
F·
∫
C
1
dr
dt
dt
(2t · t2 , t2 ) · (1, 2t)dt
=
0
=1
(b)
C1 : r = (t, 0) (0 ≤ t ≤ 1)
C2 : r = (1, t) (0 ≤ t ≤ 1)
∫
∫
F · dr =
C
F·
C1
∫ 1
dr
dt +
dt
∫
F·
C2
(b)
dr
dt
dt
C1 : r = (t, 0) (0 ≤ t ≤ 1)
C2 : r = (1, t) (0 ≤ t ≤ 1)
(3t2 , 0) · (1, 0)dt
=
∫
∫
dr
dr
dt +
F·
dt
F · dr =
F·
dt
dt
C2
C
C1
∫ 1
=
(0, t2 ) · (1, 0)dt
∫
0
1
(3 + 2t, −5t) · (0, 1)dt
+
0
=1−
5
2
∫
0
1
∫
3
=−
2
(2t, 1) · (0, 1)dt
+
0
=1
(c)
C : r = (t, t) (0 ≤ t ≤ 1)
∫
(c)
∫
C : r = (t, t) (0 ≤ t ≤ 1)
dr
dt
F · dr =
F·
dt
C
C
∫ 1
=
(3t2 + 2t, −5t2 ) · (1, 1)dt
∫
∫
F · dr =
C
0
2
=− +1
3
1
=
3
F·
C
1
∫
(2t2 , t2 ) · (1, 1)dt
=
0
=1
1
dr
dt
dt
以下、書き込み禁止
(2)
∂Fx
∂Fy
−
= 2x − 2x
∂y
∂x
=0
(3)
∂f (r)
= −2xy
∂x
∂f (r)
= −x2
∂y
(1)
(2)
(1) 式を x で積分する。
f (r) = −x2 y + C(y) (C(y) : 積分定数)
上式を y で微分する。
dC(y)
∂f (r)
= −x2 +
∂y
dy
= −x2
(∵ (2) 式)
これより
dC(y)
=0
dy
C(y) = const.
∴ f (r) = −x2 y + const.
(4)
−[f (1, 1) − f (0, 0)]
= −[−1 + C − (0 + C)]
=1
2

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